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10. Binary Tree Inorder Traversal

easyAsked at GitHub

Return the inorder traversal of a binary tree's node values — GitHub's lead-in to traversing the parent-DAG of a commit history in chronological order.

By Alex Chen, Founder, InterviewChamp.AI · Last verified

Problem

Given the root of a binary tree, return the inorder traversal of its nodes' values. Solve recursively, then convert to iterative using an explicit stack.

Constraints

  • Nodes in [0, 100]
  • -100 <= Node.val <= 100

Examples

Example 1

Input
root = [1,null,2,3]
Output
[1,3,2]

Example 2

Input
root = []
Output
[]

Approaches

1. Recursive

Standard left-root-right recursion.

Time
O(n)
Space
O(h)
function inorder(root, out=[]) {
  if (!root) return out;
  inorder(root.left, out);
  out.push(root.val);
  inorder(root.right, out);
  return out;
}

Tradeoff:

2. Iterative stack

Push lefts onto stack, pop and visit, then dive right. Avoids stack-overflow on deep trees and matches how commit-graph walkers iterate parents lazily.

Time
O(n)
Space
O(h)
function inorderTraversal(root) {
  const stack = [], out = [];
  let cur = root;
  while (cur || stack.length) {
    while (cur) { stack.push(cur); cur = cur.left; }
    cur = stack.pop();
    out.push(cur.val);
    cur = cur.right;
  }
  return out;
}

Tradeoff:

GitHub-specific tips

GitHub asks 'now do it iteratively' — they care about explicit-stack mastery because the commit-graph walker can't risk JS recursion limits on deep repos.

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Output

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